Original Message ----- Tail light Brightness
One of our listers, last night, (Friday) wrote correctly, about the tail
light being less bright. However, they were way off in the math regarding
the statement that I snipped below.
It was mentioned, that a variable power supply made a 12 volt light work
nice and normal.
Then, when he halved the voltage to 6 volts ...the light was only half as
bright.
Not true.
It's called the inverese square law... and you can't change physics.
I have entered the formula showing what happens when you "halve" the
voltage on "anything"
Halve the voltage you get only one quarter the power (not half, as
mentioned.)
In other words, whether you speaking of anything electrical in a Direct
Current - DC circuit, (which is our Mil-Vehicles - Willys vehicles etc.)
Snipped from previous post;
Imagine it this way. I have a variable voltage power supply I set
it to 12 volts and hook it up to a 12V lamp. I glows nice and normal, then
I turn the dial down to 6....half as bright...right?
I'm not perfect and make mistakes too often to admit, but on this one, if
I'm wrong, I'm in the wrong line of work :)
(end of quote)
Rob writes,
Well, don't throw in the towel just yet, I think your line of works suits
you wonderfully.!
During the 1950's comon lamp sizes / brightness were
Tail lights equaled 6 watts each
Stop lights equaled 18watts each
Head lights equaled 35watts each
(these numbers were what "Lucas" and "Smith" of England were using on their
early 12 volts systems)
To a certai degree, it doesn't matter what voltage a device is designed &
operated with ...the "Wattage" is the standard measurement for how we buy
our light bulbs nowdays for our homes.
(let's not get into Lumens - Foot candles - Joules etc, and the inefficiency
of the little filament)
I
t is the same for a toaster an electric stove and "all" other non inductive
loads
A windsheild wiper-12 Volt from a later vehicle operating on 6 volts will
barely move
Same problem with an heater fan, it will barely turn
So, by halving the volts, unfortunately you wont get half the speed !
I will take the "Stop light" 18 watt filament, as an example:
In the formulas we'll use;
P = Power - measured in Watts
I = Current - measured in amps
R = Resistance - measured in ohms
V = Volts - measured in Volts
To determine Power we multiply Volts by Current
P = E x I
Determine Current, divide Volts by Resistance
I = E / R
Determine Resistance, divide Volts by Current
R = E / I
Determine Volts, multiply Current by Resistance
E = I x R
Therefore, in our stop light bulb we have
P = E x I or 18 watts = 12 volts x amps
But we dont know the amps required, do we?
Therefore, we divide the Volts into watts.
I = P / E or I = 18 / 12
I = 1.5 amps
Now, to look at what we get when we run a 12volt lamp on 6 volts
...and this is where many of us "assume" that we will get "Half" the
brightness !
The lamp filament has an 8ohm resistance
How do we know that?
Well, we know the volts and amps ....
Therefore, R = Volts divided by amps or
R = E / I which is 12v divided by 1.5amps
Which is R = 12 / 1.5 = 8ohms
Now we know the voltage and current.
So lets see if this works for the Power formula ?
Where power in watts is P = E x I
P = 12 x 1.5
P = 18 watts
Here's the point ...and I do have one !
Let us see what happens to the Power in watts of our lowly Stop light when
we halve the voltage across it's filament.
(which is where I responded with this long posting)
Firstly, we have to find what amps would be flowing with our12volt stop
light, operating on 6 volts.
Regardless, we know that the filament is still 8ohms, That can't change.
We also know that now we only have a 6 volt battery supply.
Therfore, amps or current
I = E / R
I = 6v / 8ohms
I = .75 amp
less than one amp now flows when we halve the voltage.
Finally, we get to our answer.
If watt's P = E x I
Then, P = 6v x .75amp
Power is now "Only" 4.5 watts
it's only one quarter of our original 18 watt lightbulb's output
This also works the other way when we place a 6volt lamp across 12volts do
we get twice the brightness?
No ! Until it blows a moment later, we get 4 times the light !
Silly, eh?
See what happens when we run our 6 volt starter motors on 12 volts
Do they run twice as fast or 4 times as fast?
A 6 volt wiper motor running on 12 volts? Burnout !
Here's an often asked question on these lists
Can I run my 6volt fuel tank sender on my new 12 volts system?
You do the math on this one.
The sender is constantly changing it's resisitance, right ?
So, check the resistance at the 4 obvious conditions
1/4 full - 1/2 full - 3/4 full - full tank position
Then do the math as a 6volt sender and then on 12volts Hhhhmmm !
Next, figure the gauge itself into the equations.
Finally, the above theory of the "Inverse square law" works in other
situations also.
For instance, with light too !
Lets say that the light in the ceiling of your workshop is really too dim.
You cant get brighter bulbs. either ?
As, in one of those fixtures with two 40watt fluro's ?
Then, if you lowered the light fixture from 10 feet above your bench to
just 5 feet above your bench ...what would the brightness be?
So, Halving that distance would make the bench area twice as bright...yes ?
No !
It would be 4 times as bright ..
Silly isn't it ...eh ?
There are other losses that occur in electrical circuits that I did not get
into during this post. The metal filament changes it resistance when it is
hot as compared to when it is cold.
So, I stayed with just the basics !
Regards
Rob Pearson
Tending my flock of early jeeps in upstate, NY area.
This archive was generated by hypermail 2b29 : Sun Jan 06 2002 - 22:26:52 PST